Question

A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 86 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 18 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?

Answer 1

The frequency of the wave is 1075 Hz, and the wavelength is 0.08 m. The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave with specific parameters. The amplitude is 0.04 m, the wave number is 25π m⁻¹, the angular frequency is 6755π rad/s, and the phase angle can be either +π or -π. The choice of sign in front of ω determines the direction of wave propagation.

Step-by-step explanation:

To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the speed of the wave, and λ is the wavelength. In this case, the speed of the wave is given as 86 m/s. Since the wave is sinusoidal, the wavelength is equal to twice the amplitude of the transverse displacement of the string particle at x = 0. So, the wavelength is equal to 2 times 4.0 cm, which is 8.0 cm or 0.08 m.

Substituting these values into the formula, we get f = 86 m/s / 0.08 m = 1075 Hz. Therefore, the frequency of the wave is 1075 Hz.

To find the wavelength of the wave, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of the wave, and f is the frequency. Substituting the given values, we get λ = 86 m/s / 1075 Hz = 0.08 m.

Therefore, the wavelength of the wave is 0.08 m.

The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave. In this equation, ym is the amplitude of the wave, k is the wave number, ω is the angular frequency, and φ is the phase angle. The choice of the sign in front of ω determines the direction of wave propagation. If it is positive, the wave propagates in the positive x-direction, and if it is negative, the wave propagates in the negative x-direction.

In this case, the amplitude of the wave is given as 4.0 cm or 0.04 m. The wave number can be calculated using the formula k = 2π/λ, where k is the wave number and λ is the wavelength. Substituting the given wavelength of 0.08 m, we get k = 2π/0.08 m = 25π m⁻¹.

The angular frequency can be calculated using the formula ω = 2πf, where ω is the angular frequency and f is the frequency. Substituting the given frequency of 1075 Hz, we get ω = 2π × 1075 Hz = 6755π rad/s.

The phase angle is given as ± π. The choice between +π and -π determines the phase of the wave at x = 0 and t = 0.

Therefore, (c) ym = 0.04 m, (d) k = 25π m⁻¹, (e) ω = 6755π rad/s, (f) φ = ±π, and (g) the correct choice of sign in front of ω depends on the desired direction of wave propagation.

Answer 2

a) 71.62 Hz

b) 1.2 m

c) 0.04m

d) 5.23 1/m

e) 450 1/s

f) Ф=π/2

g) minus sign

Step-by-step explanation:

Hi!

Let's start with the wave equation and then write all the conditions that aregiven to us:

The text says that at t=0 and x=0 the transverse displacement is 4.0 cm:

`0.04=y(0,0)=y_(m)sin(\phi)` -- (1)

And also it says that the particle is not moving this means that its transversal velocity is zero:

`0m/s =(dy)/(dt)(0,0)=(+-)\omega y_(m)cos(\phi)`

For this to be zero, and not reduce to a trivial solution (y=0)

cos(Ф) = 0 ---> Ф=π/2 -- (2)

Repacing (2) in (1) we get:

`0.04=y(0,0)=y_(m)sin(\phi) = y_(m)sin(\pi /2)=y_(m)`

Therefore

[text]y_{m}=0.04[/text] --- (3)

Now it says that the maximum transverse speed at x=0 is 18m/s this means that:

ωA=18m/s

and since A=0.04m we get that

ω = 450 1/s --(4)

The relation between the frequency (f) and angular frequency(ω) is given by:

ω=2πf

Therefore:

a)

The frequency of the wave is:

f = ω/2π = 71.62 Hz

b)

To calculate the wavelength we must use the following expression between the wavelength, the frequency and the velocity of the wave:

v=fλ

so the wavelength is:

λ=v/f =1.2 m

c)

We already have calculated this value in eq (3)

ym=0.04m

d)

The relationship between the wavenumber and the wavelength is:

k=2π/λ

Therefore

k=5.23 1/m

e)

We also calculated this one before at eq(4)

ω = 450 1/s

f)

From eq(2)

Ф=π/2

g)

The correct choice of the sign of ω is the minus sign since the wave is traveling to the positive direction of the x axis